Question

The force $F_1$ that is necessary to move a body up an inclined plane is double the force $F_2$ that is necessary to just prevent it from sliding down, then: ( Given that $\varphi$ is angle of repose, $\theta$ is angle of inclined plane and $\text{w}$ is weight of the body)

• A. $F_2=\text{w sin}(\theta -\varphi )\text{sec}\varphi$
• B. $F_1=\text{w sin}(\theta -\varphi )\text{sec}\varphi$
• C. $\text{tan}\varphi =3\text{tan}\theta$
• D. $\text{tan}\theta =3\text{tan}\varphi$

Answer 1

Answer: (A), (D)

The correct options are
A $F_2=\text{w sin}(\theta -\varphi )\text{sec}\varphi$
D $\text{tan}\theta =3\text{tan}\varphi$


$F_1=\text{mg sin}\theta +\mu \text{mg cos}\theta .$
$F_2=\text{mg sin}\theta -\mu \text{mg cos}\theta .$
But $\text{mg = w}$
$\mu =\text{tan}\varphi F_1=\text{w (sin}\theta +\frac{\text{sin}\varphi }{\text{cos}\varphi }\text{cos}\theta )\Rightarrow F_1=\text{w sin}(\theta +\varphi )\text{sec}\varphi$
$F_2=\text{w (sin}\theta -\frac{\text{sin}\varphi }{\text{cos}\varphi }\text{cos}\theta )\Rightarrow F_2=\text{w sin}(\theta -\varphi )\text{sec}\varphi$
Now $F_1=2F_2$
$\text{mg sin}\theta +\mu \text{mg cos}\theta =2(\text{mg sin}\theta -\mu \text{mg cos}\theta )$
$\text{sin}\theta +\mu \text{cos}\theta =2\text{sin}\theta -2\mu \text{cos}\theta \Rightarrow 3\mu \text{cos}\theta =\text{sin}\theta$
$\Rightarrow \text{tan}\theta =3\mu$

$\therefore \text{tan}\theta =3\text{tan}\varphi .$